Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $n = \dfrac{x - 9}{x^2 + 9x - 10} \times \dfrac{x - 1}{2x - 18} $
Answer: First factor the quadratic. $n = \dfrac{x - 9}{(x - 1)(x + 10)} \times \dfrac{x - 1}{2x - 18} $ Then factor out any other terms. $n = \dfrac{x - 9}{(x - 1)(x + 10)} \times \dfrac{x - 1}{2(x - 9)} $ Then multiply the two numerators and multiply the two denominators. $n = \dfrac{ (x - 9) \times (x - 1) } { (x - 1)(x + 10) \times 2(x - 9) } $ $n = \dfrac{ (x - 9)(x - 1)}{ 2(x - 1)(x + 10)(x - 9)} $ Notice that $(x - 9)$ and $(x - 1)$ appear in both the numerator and denominator so we can cancel them. $n = \dfrac{ (x - 9)\cancel{(x - 1)}}{ 2\cancel{(x - 1)}(x + 10)(x - 9)} $ We are dividing by $x - 1$ , so $x - 1 \neq 0$ Therefore, $x \neq 1$ $n = \dfrac{ \cancel{(x - 9)}\cancel{(x - 1)}}{ 2\cancel{(x - 1)}(x + 10)\cancel{(x - 9)}} $ We are dividing by $x - 9$ , so $x - 9 \neq 0$ Therefore, $x \neq 9$ $n = \dfrac{1}{2(x + 10)} ; \space x \neq 1 ; \space x \neq 9 $